JavaScript Spiral Matrix Coding Challenge












0












$begingroup$


A whiteboarding challenge: Given a 2D array (matrix) inputMatrix of integers, create a function spiralCopy that copies inputMatrix’s values into a 1D array in a clockwise spiral order. Your function then should return that array.



inputMatrix = [

  [1, 2, 3, 4, 5, 6],

  [7, 8, 9, 10, 11, 12],

  [13, 14, 15, 16, 17, 18],

  [19, 20, 21, 22, 23, 24],

  [25, 26, 27, 28, 29, 30],

  [31, 32, 33, 34, 35, 36]

]



const spiralCopy = (matrix) => {

  const spiralLength = matrix.length * matrix[0].length;

  const spiral = ;



  let iLeftRightStart = 0; // +1

  let iLeftRightEnd = matrix[0].length; // -1

  let leftRightPosition = 0; // +1

  let iTopBottomStart = 1; // +1

  let iTopBottomEnd = matrix.length; // -1

  let topBottomPosition = matrix[0].length - 1; // -1

  let iRightLeftStart = matrix[0].length - 2; // -1

  let iRightLeftEnd = 0; // +1

  let rightLeftPosition = matrix.length - 1; // -1

  let iBottomTopStart = matrix.length - 2; // -1

  let iBottomTopEnd = 1; // +1                      

  let bottomTopPosition = 0; // +1



  const leftToRight = (iStart, iEnd, position) => {

    for (let i = iStart; i < iEnd; i++) {

      spiral.push(matrix[position][i]);

    }

  }



  const topToBottom = (iStart, iEnd, position) => {

    for (let i = iStart; i < iEnd; i++) {

      spiral.push(matrix[i][position]);

    }

  }



  const rightToLeft = (iStart, iEnd, position) => {

    for (let i = iStart; i >= iEnd; i--) {

      spiral.push(matrix[position][i]);

    }

  }



  const bottomToTop = (iStart, iEnd, position) => {

    for (let i = iStart; i >= iEnd; i--) {

      spiral.push(matrix[i][position]);

    }

  }



  while (spiral.length < spiralLength) {

    leftToRight(iLeftRightStart, iLeftRightEnd, leftRightPosition);

    topToBottom(iTopBottomStart, iTopBottomEnd, topBottomPosition);

    rightToLeft(iRightLeftStart, iRightLeftEnd, rightLeftPosition);

    bottomToTop(iBottomTopStart, iBottomTopEnd, bottomTopPosition);

    iLeftRightStart++;

    iLeftRightEnd--;

    leftRightPosition++;

    iTopBottomStart++;

    iTopBottomEnd--;

    topBottomPosition--;

    iRightLeftStart--;

    iRightLeftEnd++;

    rightLeftPosition--;

    iBottomTopStart--;

    iBottomTopEnd++;

    bottomTopPosition++;

  }

  return spiral;

}

console.log(spiralCopy(inputMatrix));

// prints [1, 2, 3, 4, 5, 6, 12, 18, 24, 30, 36, 35, 34, 33, 32, 31, 25, 19, 13, 7, 8, 9, 10, 11, 17, 23, 29, 28, 27, 26, 20, 14, 15, 16, 22, 21]


So the function works and runs with time complexity of O(N). However, I don't like that I'm using 12 different variables which I then increment/decrement. How can this be consolidated, and is it possible to make the algorithm even faster due to the instant access of JavaScript arrays?






javascript programming-challenge array matrix







share|improve this question









$endgroup$

















    0












    $begingroup$


    A whiteboarding challenge: Given a 2D array (matrix) inputMatrix of integers, create a function spiralCopy that copies inputMatrix’s values into a 1D array in a clockwise spiral order. Your function then should return that array.



    inputMatrix = [
    
  [1, 2, 3, 4, 5, 6],
    
  [7, 8, 9, 10, 11, 12],
    
  [13, 14, 15, 16, 17, 18],
    
  [19, 20, 21, 22, 23, 24],
    
  [25, 26, 27, 28, 29, 30],
    
  [31, 32, 33, 34, 35, 36]
    
]
    

    
const spiralCopy = (matrix) => {
    
  const spiralLength = matrix.length * matrix[0].length;
    
  const spiral = ;
    

    
  let iLeftRightStart = 0; // +1
    
  let iLeftRightEnd = matrix[0].length; // -1
    
  let leftRightPosition = 0; // +1
    
  let iTopBottomStart = 1; // +1
    
  let iTopBottomEnd = matrix.length; // -1
    
  let topBottomPosition = matrix[0].length - 1; // -1
    
  let iRightLeftStart = matrix[0].length - 2; // -1
    
  let iRightLeftEnd = 0; // +1
    
  let rightLeftPosition = matrix.length - 1; // -1
    
  let iBottomTopStart = matrix.length - 2; // -1
    
  let iBottomTopEnd = 1; // +1                      
    
  let bottomTopPosition = 0; // +1
    

    
  const leftToRight = (iStart, iEnd, position) => {
    
    for (let i = iStart; i < iEnd; i++) {
    
      spiral.push(matrix[position][i]);
    
    }
    
  }
    

    
  const topToBottom = (iStart, iEnd, position) => {
    
    for (let i = iStart; i < iEnd; i++) {
    
      spiral.push(matrix[i][position]);
    
    }
    
  }
    

    
  const rightToLeft = (iStart, iEnd, position) => {
    
    for (let i = iStart; i >= iEnd; i--) {
    
      spiral.push(matrix[position][i]);
    
    }
    
  }
    

    
  const bottomToTop = (iStart, iEnd, position) => {
    
    for (let i = iStart; i >= iEnd; i--) {
    
      spiral.push(matrix[i][position]);
    
    }
    
  }
    

    
  while (spiral.length < spiralLength) {
    
    leftToRight(iLeftRightStart, iLeftRightEnd, leftRightPosition);
    
    topToBottom(iTopBottomStart, iTopBottomEnd, topBottomPosition);
    
    rightToLeft(iRightLeftStart, iRightLeftEnd, rightLeftPosition);
    
    bottomToTop(iBottomTopStart, iBottomTopEnd, bottomTopPosition);
    
    iLeftRightStart++;
    
    iLeftRightEnd--;
    
    leftRightPosition++;
    
    iTopBottomStart++;
    
    iTopBottomEnd--;
    
    topBottomPosition--;
    
    iRightLeftStart--;
    
    iRightLeftEnd++;
    
    rightLeftPosition--;
    
    iBottomTopStart--;
    
    iBottomTopEnd++;
    
    bottomTopPosition++;
    
  }
    
  return spiral;
    
}
    
console.log(spiralCopy(inputMatrix));
    
// prints [1, 2, 3, 4, 5, 6, 12, 18, 24, 30, 36, 35, 34, 33, 32, 31, 25, 19, 13, 7, 8, 9, 10, 11, 17, 23, 29, 28, 27, 26, 20, 14, 15, 16, 22, 21]


    So the function works and runs with time complexity of O(N). However, I don't like that I'm using 12 different variables which I then increment/decrement. How can this be consolidated, and is it possible to make the algorithm even faster due to the instant access of JavaScript arrays?






    javascript programming-challenge array matrix







    share|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      A whiteboarding challenge: Given a 2D array (matrix) inputMatrix of integers, create a function spiralCopy that copies inputMatrix’s values into a 1D array in a clockwise spiral order. Your function then should return that array.



      inputMatrix = [
      
  [1, 2, 3, 4, 5, 6],
      
  [7, 8, 9, 10, 11, 12],
      
  [13, 14, 15, 16, 17, 18],
      
  [19, 20, 21, 22, 23, 24],
      
  [25, 26, 27, 28, 29, 30],
      
  [31, 32, 33, 34, 35, 36]
      
]
      

      
const spiralCopy = (matrix) => {
      
  const spiralLength = matrix.length * matrix[0].length;
      
  const spiral = ;
      

      
  let iLeftRightStart = 0; // +1
      
  let iLeftRightEnd = matrix[0].length; // -1
      
  let leftRightPosition = 0; // +1
      
  let iTopBottomStart = 1; // +1
      
  let iTopBottomEnd = matrix.length; // -1
      
  let topBottomPosition = matrix[0].length - 1; // -1
      
  let iRightLeftStart = matrix[0].length - 2; // -1
      
  let iRightLeftEnd = 0; // +1
      
  let rightLeftPosition = matrix.length - 1; // -1
      
  let iBottomTopStart = matrix.length - 2; // -1
      
  let iBottomTopEnd = 1; // +1                      
      
  let bottomTopPosition = 0; // +1
      

      
  const leftToRight = (iStart, iEnd, position) => {
      
    for (let i = iStart; i < iEnd; i++) {
      
      spiral.push(matrix[position][i]);
      
    }
      
  }
      

      
  const topToBottom = (iStart, iEnd, position) => {
      
    for (let i = iStart; i < iEnd; i++) {
      
      spiral.push(matrix[i][position]);
      
    }
      
  }
      

      
  const rightToLeft = (iStart, iEnd, position) => {
      
    for (let i = iStart; i >= iEnd; i--) {
      
      spiral.push(matrix[position][i]);
      
    }
      
  }
      

      
  const bottomToTop = (iStart, iEnd, position) => {
      
    for (let i = iStart; i >= iEnd; i--) {
      
      spiral.push(matrix[i][position]);
      
    }
      
  }
      

      
  while (spiral.length < spiralLength) {
      
    leftToRight(iLeftRightStart, iLeftRightEnd, leftRightPosition);
      
    topToBottom(iTopBottomStart, iTopBottomEnd, topBottomPosition);
      
    rightToLeft(iRightLeftStart, iRightLeftEnd, rightLeftPosition);
      
    bottomToTop(iBottomTopStart, iBottomTopEnd, bottomTopPosition);
      
    iLeftRightStart++;
      
    iLeftRightEnd--;
      
    leftRightPosition++;
      
    iTopBottomStart++;
      
    iTopBottomEnd--;
      
    topBottomPosition--;
      
    iRightLeftStart--;
      
    iRightLeftEnd++;
      
    rightLeftPosition--;
      
    iBottomTopStart--;
      
    iBottomTopEnd++;
      
    bottomTopPosition++;
      
  }
      
  return spiral;
      
}
      
console.log(spiralCopy(inputMatrix));
      
// prints [1, 2, 3, 4, 5, 6, 12, 18, 24, 30, 36, 35, 34, 33, 32, 31, 25, 19, 13, 7, 8, 9, 10, 11, 17, 23, 29, 28, 27, 26, 20, 14, 15, 16, 22, 21]


      So the function works and runs with time complexity of O(N). However, I don't like that I'm using 12 different variables which I then increment/decrement. How can this be consolidated, and is it possible to make the algorithm even faster due to the instant access of JavaScript arrays?






      javascript programming-challenge array matrix







      share|improve this question









      $endgroup$




      A whiteboarding challenge: Given a 2D array (matrix) inputMatrix of integers, create a function spiralCopy that copies inputMatrix’s values into a 1D array in a clockwise spiral order. Your function then should return that array.



      inputMatrix = [
      
  [1, 2, 3, 4, 5, 6],
      
  [7, 8, 9, 10, 11, 12],
      
  [13, 14, 15, 16, 17, 18],
      
  [19, 20, 21, 22, 23, 24],
      
  [25, 26, 27, 28, 29, 30],
      
  [31, 32, 33, 34, 35, 36]
      
]
      

      
const spiralCopy = (matrix) => {
      
  const spiralLength = matrix.length * matrix[0].length;
      
  const spiral = ;
      

      
  let iLeftRightStart = 0; // +1
      
  let iLeftRightEnd = matrix[0].length; // -1
      
  let leftRightPosition = 0; // +1
      
  let iTopBottomStart = 1; // +1
      
  let iTopBottomEnd = matrix.length; // -1
      
  let topBottomPosition = matrix[0].length - 1; // -1
      
  let iRightLeftStart = matrix[0].length - 2; // -1
      
  let iRightLeftEnd = 0; // +1
      
  let rightLeftPosition = matrix.length - 1; // -1
      
  let iBottomTopStart = matrix.length - 2; // -1
      
  let iBottomTopEnd = 1; // +1                      
      
  let bottomTopPosition = 0; // +1
      

      
  const leftToRight = (iStart, iEnd, position) => {
      
    for (let i = iStart; i < iEnd; i++) {
      
      spiral.push(matrix[position][i]);
      
    }
      
  }
      

      
  const topToBottom = (iStart, iEnd, position) => {
      
    for (let i = iStart; i < iEnd; i++) {
      
      spiral.push(matrix[i][position]);
      
    }
      
  }
      

      
  const rightToLeft = (iStart, iEnd, position) => {
      
    for (let i = iStart; i >= iEnd; i--) {
      
      spiral.push(matrix[position][i]);
      
    }
      
  }
      

      
  const bottomToTop = (iStart, iEnd, position) => {
      
    for (let i = iStart; i >= iEnd; i--) {
      
      spiral.push(matrix[i][position]);
      
    }
      
  }
      

      
  while (spiral.length < spiralLength) {
      
    leftToRight(iLeftRightStart, iLeftRightEnd, leftRightPosition);
      
    topToBottom(iTopBottomStart, iTopBottomEnd, topBottomPosition);
      
    rightToLeft(iRightLeftStart, iRightLeftEnd, rightLeftPosition);
      
    bottomToTop(iBottomTopStart, iBottomTopEnd, bottomTopPosition);
      
    iLeftRightStart++;
      
    iLeftRightEnd--;
      
    leftRightPosition++;
      
    iTopBottomStart++;
      
    iTopBottomEnd--;
      
    topBottomPosition--;
      
    iRightLeftStart--;
      
    iRightLeftEnd++;
      
    rightLeftPosition--;
      
    iBottomTopStart--;
      
    iBottomTopEnd++;
      
    bottomTopPosition++;
      
  }
      
  return spiral;
      
}
      
console.log(spiralCopy(inputMatrix));
      
// prints [1, 2, 3, 4, 5, 6, 12, 18, 24, 30, 36, 35, 34, 33, 32, 31, 25, 19, 13, 7, 8, 9, 10, 11, 17, 23, 29, 28, 27, 26, 20, 14, 15, 16, 22, 21]


      So the function works and runs with time complexity of O(N). However, I don't like that I'm using 12 different variables which I then increment/decrement. How can this be consolidated, and is it possible to make the algorithm even faster due to the instant access of JavaScript arrays?






      javascript programming-challenge array matrix




      javascript programming-challenge array matrix






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 58 mins ago









      Sean ValdiviaSean Valdivia

      3751414




      3751414






















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